Question

In which one of the following reactions, $+2$ state in a reactant is reduced to $+1$ state?

• A. $Mg + S \rightarrow MgS$
• B. $2HgCl_2 + SnCl_2 \rightarrow Hg_2Cl_2 + SnCl_4$
• C. $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$
• D. $2K_4[Fe(CN)_6] + H_2O_2 \rightarrow 2K_3[Fe(CN)_6] + 2KOH$

Hint:

Identify the metal ions first. Mercury is unique as it forms both $Hg^{2+}$ (mercuric) and $Hg_2^{2+}$ (mercurous) ions, where the latter has an oxidation state of $+1$ per atom.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Reduction involves the decrease in the oxidation state of an element. We need to find a reaction where a reactant atom starting at +2 ends at +1.
Step 2: Key Formula or Approach:
Calculate the oxidation state of key elements in each reaction.
Oxidation state of Cl is usually -1.
Step 3: Detailed Explanation:
Analyze Option B: \( 2\text{HgCl}_2 + \text{SnCl}_2 \rightarrow \text{Hg}_2\text{Cl}_2 + \text{SnCl}_4 \)
- In HgCl\textsubscript{2}, Hg is in the +2 state.
- In Hg\textsubscript{2}Cl\textsubscript{2} (calomel), each Hg is in the +1 state.
- In SnCl\textsubscript{2}, Sn is +2; in SnCl\textsubscript{4}, Sn is +4 (Oxidation).
Thus, Mercury (Hg) is reduced from +2 to +1.
Other options:
A: Mg(0) \(\rightarrow\) Mg(+2).
D: Fe(+2) \(\rightarrow\) Fe(+3).
Step 4: Final Answer:
Reaction B shows the reduction of +2 state to +1.